the meaning of "const" in C/C++ with functions

koyi · 09-12-2006, 05:38 AM

Hi, I have a simple(?) question about the type-qualifier "const" in use with functions declaration...

What does this mean?

Code:

const char function_a () {...

And what about this?

Code:

char function_b () const { ...

Thanks

Guttorm · 09-12-2006, 08:46 AM

Hi

In C, the "const" prefix is for any type - and is used only to generate an error (or warning) if the value changes after an assignment.

E.g.
const char c;
c = 'a';
c = 'b';

It is mainly used for pointers, so the compiler can prevent mistakes like this
const char * function_a()
{
return "a";
}

char *ptr = function_a();
strcpy(ptr,"b");

The other use, is telling the compiler that the function is returning something based on its parameters only and the result will be the same every time you call it with the same parameters.

So for example the function:
int fact(int n) const
{
int i, sum;
for ( sum=1,i=2 ; i<=n ; i++ )
sum *= i;
return sum;
}

if you later use something like
int x = fact(32);
the compiler can optimize and there will be no loop executed in the program - it can calculate the number at compile time.

Not all compilers support this use of "const" though.

mjones490 · 09-12-2006, 09:54 AM

Also, in the second use (putting const after the function signature and before the function body), when used in a member function of a class, makes the function promise not to alter the object of the class, and this function can be called from a constant instance. Like this:

Code:

class cApplesauce
{
private:
   int m_iTheValue;

public:
   void setTheValue (int iNewValue)
   {
       m_iTheValue = iNewValue;
   }

   int getTheValue () const
   {
       return m_iTheValue;
   }
}

This example class just has set/get functions for m_iTheValue. Notice const after the getTheValue function.

Code:

cApplesauce theObject;
theObject.setTheValue (42);
cout << theObject.getTheValue () << endl;

const cApplesauce theConstObject = theObject;
theConstObject.setTheValue (69); // Not allowed!
cout << theConstObject.getTheValue () << endl;

This code snippet instantiates a regular object, sets its value, and prints it. It is okay to set the value.

Then it instantiates a const object (by copying the regular object (which will set its value to 42)), and attempts to set its value to 69. That won't work; the set function has not promised to change the value, and this is a const object. Then it prints the value, which works because the get function HAS promised not to change anything.

Thanks,
Mark

koyi · 09-13-2006, 04:40 AM

Wow... you guys really enlightened me... din know the simple const has so many meanings...

Thanks

Abomb · 09-13-2006, 05:13 PM

Just to add a little note on using const with pointers, which is a good idea. You can define a pointer numerous ways:

Code:

char* ptr; // Non-const pointer.
const char* cptr; // Promise not to change the pointer itself.
char* const ptrc; // Promise not to change what the pointer is pointing to.
const char* const cptrc; // Promise not to change the pointer AND what it's pointing to.

This might not look like it's important but:

Code:

++ptr;    // legal.
++cptr;   // legal.
++ptrc;   // illegal.
++cptrc;  // illegal.